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What type of singularity is that at $z=\pi k+\pi/2$ for any integer $k$ for the function $\phi(z)=e^{\tan z}$? I can see that it is not removable, but I am not sure how to narrow down further. Thank you.

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It's a bit hard to give the proper hints without giving the whole answer away, especially not knowing what you already know about the classification of singularities. Briefly, though: What do you already know about the singularities of $\tan z$ at these points, and are you familiar with the singularity of $e^{-1/z}$ at $z=0$? – Harald Hanche-Olsen Feb 19 '12 at 21:53
    
@HaraldHanche-Olsen: I am guessing that they are simple poles (so order 1)? But I am not familiar with that of $e^{-1/z}$ at $z=0$. Though if I had to guess, I think it is essential? – Leon C Feb 19 '12 at 22:09
    
You're right on both counts. If $e^{-1/z}$ had a pole at the origin, there should be some natural number $n$ so that $z^ne^{-1/z}$ had a removable singularity there. Can you see how that cannot possibly be true? Does it help with your original question? – Harald Hanche-Olsen Feb 19 '12 at 22:20
    
@HaraldHanche-Olsen: Because the Taylor series of $e^{-1/z}$ has terms in $z^n$ for all natural numbers $n$? – Leon C Feb 19 '12 at 23:19
    
Yes, that is one way to see it. Another is to note that $z^ne^{-1/z}$ is unbounded in every neighbourhood of the origin. – Harald Hanche-Olsen Feb 20 '12 at 6:36

When a meromorphic function $f$ has a pole at $a$, the composition $e^f$ has an essential singularity at $a$. One way to see this is: neither $e^f$ nor $1/e^f=e^{-f}$ are bounded in any neighborhood of $a$. (Pointed out by Harald Hanche-Olsen).

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